Question: Solve the equation \[-x^2 = \frac{3x+1}{x+3}.\]Enter all solutions, separated by commas.
Explanation: Multiplying both sides by $x+3,$ we have $-x^2(x+3) = 3x+1,$ or $-x^3 - 3x^2 = 3x + 1.$ Thus, \[x^3 + 3x^2 + 3x + 1 = 0.\]We recognize the left-hand side as the expansion of $(x+1)^3,$ so \[(x+1)^3 = 0.\]This forces $x+1=0,$ so $x = \boxed{-1},$ which is the only solution.